Question #10e47

2 Answers
Jan 18, 2018

Answer:

Toque applied = #6 N.m# & power is about#120# watt

Explanation:

Here,by applying the torque angular retardation was done to stop its motion(considering no skidding was there).

so, angular retardation= #alpha# = change in angular velocity i.e #omega/t# = #40/10# i.e #4# radian/#s^2#

Now, torque =# Ialpha# (where, I is the moment of inertia)

moment of inertia of a cylinder passing through its axis is #(Mr^2)/2# =#1.5 Kg.m^2#

so. #tau# =# 1.5*4 N.m# i,e #6 N.m#

so,angular power required = #(tau* theta)/t# i.e about #120# watt(#tau*theta#= work done,dividing by time required for this angular displacement of #theta# by the time #t# in which it was brought to rest due to application of this torque) ( #theta# is angular displacement)(# theta = (omega)^2/(2 alpha)=(40^2)/(2*4)=200 radian#)

Jan 18, 2018

Answer:

The torque is #=6Nm#. The power is #=240W#

Explanation:

The torque and the angular acceleration are related

#"Torque (Nm)"="Moment of inertia (kgm^2)"xx "angular acceleration (rads^-2)"#

#tau=Ixxalpha#

The moment of inertia of the solid cylinder is

#I=(mr^2)/2#

The mass of the cylinder is #m=3kg#

The radius of the cylinder is #r=1.0m#

The moment of inertia is #I=(3xx1^2)/2=1.5kgm^2#

The initial angular velocity is #omega_0=40rads^-1#

The final angular velocity is #omega_1=0rads^-1#

The time is #t=10s#

Apply the equation (to find the angular acceleration #(alpha)#)

#omega_1=omega_0+alphat#

#alpha=(omega_1-omega_0)/t=(0-40)/10=-4rads^-2#

The torque to be applied is

#tau=Ialpha=1.5xx4=6Nm#

#"Power (W)"= "torque (Nm)"xx "angular velocity (rads^-1)"#

#P=tauxx omega_0#

The power is #P=6xx40=240W#